Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Example 33 Find intervals in which the function given by f(đĽ) =3/10 đĽ4 â 4/5 đĽ^3â 3đĽ2 + 36/5 đĽ + 11 is (a) strictly increasing (b) strictly decreasingf(đĽ) = 3/10 đĽ4 â 4/5 đĽ^3â 3đĽ2 + 36/5 đĽ + 11 Finding fâ(đ) fâ(đĽ) = 3/10 Ă 4đĽ^3 â 4/5 Ă 3đĽ^2 â 3 Ă 2x + 36/5 + 0 fâ(đĽ) = 12/10 đĽ^3â 12/5 đĽ^2â 6x + 36/5 fâ(đĽ) = 6/5 đĽ^3â 12/5 đĽ^2â 6x + 36/5 fâ(đĽ) = 6(đĽ^3/5â(2đĽ^2)/5âđĽ+6/5) fâ(đĽ) = 6((đĽ^3 â 2đĽ^2â 5đĽ + 6)/5) = 6/5 (đĽ^3â2đĽ^2â5đĽ+6) = 6/5 (đĽâ1)(đĽ2âđĽâ6) = 6/5 (đĽâ1)(đĽ2â3đĽ+2đĽâ6) = 6/5 (đĽâ1)[đĽ(đĽâ3)+2(đĽâ3)] = 6/5 (đĽâ1)(đĽ+2)(đĽâ3) Hence, fâ(đ) = đ/đ (đâđ)(đ+đ)(đâđ) Putting fâ(đ) = 0 đ/đ (đâđ)(đ+đ)(đâđ) = 0 fâ(đĽ) = 6((đĽ^3 â 2đĽ^2â 5đĽ + 6)/5) = 6/5 (đĽ^3â2đĽ^2â5đĽ+6) = 6/5 (đĽâ1)(đĽ2âđĽâ6) = 6/5 (đĽâ1)(đĽ2â3đĽ+2đĽâ6) = 6/5 (đĽâ1)[đĽ(đĽâ3)+2(đĽâ3)] = 6/5 (đĽâ1)(đĽ+2)(đĽâ3) Hence, fâ(đ) = đ/đ (đâđ)(đ+đ)(đâđ) Putting fâ(đ) = 0 đ/đ (đâđ)(đ+đ)(đâđ) = 0 (đĽâ1)(đĽ+2)(đĽâ3) = 0 Hence, x = â2 , 1 & 3 Plotting points on number line Hence, f(đĽ) is strictly decreasing on the interval đĽ â (ââ,âđ)& (đ , đ) f(đĽ) is strictly increasing on the interval đĽ â (âđ,đ) & (đ , â)